typescript - Tuple with a rest element which is a union type - Stack Overflow

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This works:

type number_ = number | null;

let myTuple: [string, ...number_[]];

myTuple = ['foo', 1, 2, 3];
myTuple = ['foo', 1, null, 3];

const [a, ...b] = myTuple;
// Type checker knows a is string, and b is array of number | null

Is there any way to express that without using number_?

This works:

type number_ = number | null;

let myTuple: [string, ...number_[]];

myTuple = ['foo', 1, 2, 3];
myTuple = ['foo', 1, null, 3];

const [a, ...b] = myTuple;
// Type checker knows a is string, and b is array of number | null

Is there any way to express that without using number_?

• typescript

Share Follow edited Mar 6 at 20:50 jonrsharpe 122k3030 gold badges268268 silver badges476476 bronze badges asked Mar 6 at 20:40 davetapleydavetapley 18k1515 gold badges6464 silver badges9292 bronze badges 1

• "without using number_"? Sure, like this. But do you really just mean you want to avoid a named type? Could you edit to clarify what you mean here? – jcalz Commented Mar 6 at 20:42

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1 Answer 1

Sorted by: Reset to default 1

You can avoid creating a special type by declaring the union inside parenthesis before the array symbol:

let myTuple: [string, ...(number | null)[]];

Doing it this way will still cause the variable b from your example to be typed as (number | null)[].

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